Division by zero using shell - shell

Question

I'm writing a shell program using while commands to prevent division by zero and to exit the program when the user types 99. Everytime i run it i get an error for the input of the second number which in turn does not give me the answer. Below is my attempt at this problem:
#!/bin/bash
firstNum=0
secondNum=0
answer=0
while true firstNum != 99, secondNum != 99 do
read -p "Enter first number" firstNum
read -p "Enter second number" secondNum
echo "first num $firstNum"
echo "second Num $secondNum"
if ["$secondNum" = "0"]
then
exit 1
else
echo "first number / second Number = $((firstNum/secondNum))"
echo "Answer = $answer"
fi
do
exit
done
And the following is the error message that i get
./example.sh: 10: ./example.sh: [2: not found
Any help at all is greatly appreciated.
Thanks

Answer 1

Looks like a few typos.
Replace this:
if ["$secondNum" = "0"] then
exit 1
then
echo "1st number / 2nd number = $((firstNum/secondNum))"
echo "Answer = $answer"
fi
with this:
if [ "$secondNum" = "0" ]
then
exit 1
else
echo "1st number / 2nd number = $((firstNum/secondNum))"
echo "Answer = $answer"
fi
Bash is finicky. Ensure then is on the next line and not same line as the if, you also had then instead of else on your 3rd line.

Answer 2

Your code can be simplified
while true; do
read -p "Enter first number: " firstNum
[[ "$firstNum" -eq 99 ]] && break
read -p "Enter second number: " secondNum
case "$secondNum" in
99) break ;;
0) echo "div by zero, try again"; continue ;;
esac
echo "first number / second Number = $(bc -l <<< "$firstNum/$secondNum")"
done
I'm calling out to bc to get floating point arithmetic. With the shell you're limited to integers only.
<<< is a "here-string", a string to pass to bc's stdin

Answer 3

Your code looks pretty good to me, but I changed the POSIX shell test around a bit to test for values less than 99 instead of 99 itself. And there was a stray do near the bottom of the script which I removed. Bash has [[]] style test syntax too which you may want to switch to at some point.
#!/bin/bash
firstNum=0
secondNum=0
answer=0
while [ "$firstNum" -lt 99 -a "$secondNum" -lt 99 ]
do
read -p "Enter first number " firstNum
read -p "Enter second number " secondNum
printf "first num $firstNum\n"
printf "second Num $secondNum\n"
if [ "$secondNum" = 0 ]
then
printf "$secondNum was a zero, exiting!\n"
exit 1
else
printf "first num / second Num = \$(($firstNum / $secondNum))\n"
answer=$(( firstNum / secondNum))
printf "Answer = $answer\n"
fi
done
If you are interested in learning the Bash test syntax [[]] (it's a bit more powerful than POSIX shell's test(1) []) here is an example which uses it:
#!/bin/bash
firstNum=0
secondNum=0
answer=0
while [[ "$firstNum" < 99 && "$secondNum" < 99 ]]
do
read -p "Enter first number " firstNum
read -p "Enter second number " secondNum
printf "first num $firstNum\n"
printf "second Num $secondNum\n"
if [[ "$secondNum" == 0 ]]
then
printf "$secondNum was a zero, exiting!\n"
exit 1
else
printf "first num / second Num = \$(($firstNum / $secondNum))\n"
answer=$(( firstNum / secondNum))
printf "Answer = $answer\n"
fi
done
It may also be advisable to move the tests into the body of the while loop and instead use the convention 'while [[ 1 ]]' to create a continuous loop. Here is an example doing that (using bash test syntax [[]]) and using bc so the script supports floating point arithmetic:
#!/bin/bash
firstNum=0
secondNum=0
answer=0
while [[ 1 ]]
do
read -p "Enter first number " firstNum
read -p "Enter second number " secondNum
printf "first num $firstNum\n"
printf "second Num $secondNum\n"
case "$firstNum" in
99) printf "Invalid values\n"
exit 1
;;
esac
case "$secondNum" in
99) printf "Invalid values\n"
exit 1
;;
0) printf "Invalid values\n"
exit 1
;;
esac
answer=$(printf "$firstNum / $secondNum\n" | bc -l)
printf "Floating point supported answer using bc: = $answer \n"
done